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Problems
Section 2-2 Engineering and Linear Models
P2.2-1
The element is not linear. For example, doubling the current from 2 A to 4 A does not double the voltage. Hence, the property of homogeneity is not satisfied.
P2.2-2
(a) The data points do indeed lie on a straight line. The slope of the line is 0.12 V/A and the line passes through the origin so the equation of the line is 0.12v=. The element is indeed linear.
(b) When i = 40 mA, v = (0.12 V/A)×(40 mA) = (0.12 V/A)×(0.04 A) = 4.8 mV
(c) When v = 4 V, 433.3A0.12i==
P2.2-3
(a) The data points do indeed lie on a straight line. The slope of the line is 256.5 V/A and the line passes through the origin so the equation of the line is 256.5v=. The element is indeed linear.
(b) When i = 4 mA, v = (256.5 V/A)×(4 mA) = (256.5 V/A)×(0.004 A) = 1.026 V
(c) When v = 12 V, 120.04678256.5i== A = 46.78 mA.
P2.2-4
Let i = 1 A , then v = 3i + 5 = 8 V. Next 2i = 2A but 16 = 2v ≠ 3(2i) + 5 = 11.. Hence, the property of homogeneity is not satisfied. The element is not linear.
P2.2-5
(a) 0.4 3.2 V10408vvvv=+=⇒= 0.08 A40vi==
(b) 220.4 0.801025vvvv=+⇒+−=
Using the quadratic formula 0.21.80.8, 1.0 V2v−±==−
When v = 0.8 V then 20.80.32 A2i==.